\newproblem{lay:1_5_21}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.5.21}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Determine the parametric equation of the line $M$ through $\mathbf{p}=(3,-3)$ and $\mathbf{q}=(4,1)$.
}
{
  % Solution
	In the following figure, we show how the direction vector of the requested line is $\mathbf{q}-\mathbf{p}$ (or $\mathbf{p}-\mathbf{q}$).
	\begin{center}
		\includegraphics[scale=0.5]{Tema2/lay_1_5_21.jpg}
	\end{center}
	The requested line can be expressed as a function of a free parameter $t\in\mathbb{R}$
	\begin{center}
		$\mathbf{l}(t)=\mathbf{p}+t(\mathbf{q}-\mathbf{p})=\begin{pmatrix}3\\-3\end{pmatrix}+t\left(\begin{pmatrix}4\\1\end{pmatrix}-\begin{pmatrix}3\\-3\end{pmatrix}\right)=
		   \begin{pmatrix}3\\-3\end{pmatrix}+t\begin{pmatrix}1\\4\end{pmatrix}$
	\end{center}
	Note that this line is at $\mathbf{p}$ for $t=0$ and at $\mathbf{q}$ for $t=1$.
}
\useproblem{lay:1_5_21}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
